I’ve explained what ideals are, but just like they generalize the concept of numbers, it makes sense to generalize the concept of fractions, in the form of fractional ideals.

Here is where I have to be a little more axiomatic. We’re interested in integral domains, which are defined by the six axioms of rings, plus the axioms that multiplication is commutative, there’s an element 1 such that 1**a* = *a* for all *a*, and the product of two nonzero elements is again nonzero. If R is an integral domain, we define the fraction field K of R to be the set of all fractions r1/r2, with the usual notion that r1/r2 = r3/r4 if r1r4 = r2r3, and the usual method of addition and multiplication. The main advantage of K is that it obeys a stronger axiom than the *ab* = 0 –> *a* = 0 or *b* = 0 axiom: in K, for every nonzero element *a*, we can find an element 1/*a* such that *a*(1/*a*) = 1.

Now, recall that an ideal I of R is identified by two properties: if *a* and *b* are in I, and *r* is any element of R, then *a* + *b* and *ra* are in I. So we can apply that rule to every subset of K. For example, if R is the ring of integers **Z**, then K is the field of rational numbers **Q**; (2) is an (integral) ideal of R, and (1/2) = {…-1.5, -1, -0.5, 0, 0.5, 1, 1.5…} is a fractional ideal of R.

Actually, to make it work, we need a third condition, namely that there exists an element *s* in R such that *s*I is contained in R, i.e. is an integral ideal. That takes care of annoying cases like (1/2, 1/4, 1/8, 1/16…).

As with integral ideals, if R is a principal ideal domain, then multiplying fractional ideals is just like multiplying integral ideals. We have (1/2)(2) = (1), (2/3)(1/10)(45) = (3), etc.

The most nifty thing about working with fractional ideals is that every ideal is invertible in a way. In particular, if I is any fractional ideal of R, then there exists another ideal, call it J, such that IJ = (1), as long as R is a Dedekind domain.

To prove that J exists, first note that by definition, there exists an element *s* of R such that *s*I is contained in R. Now, if *r* is any element of R, then clearly *rs*I is contained in R, and if *t* is another element of K such that *t*I is contained in R, then (*s*+*t*)I is contained in R, since it’s a subset of the larger integral ideal *s*I + *t*I. So the set of elements *s* of K such that *s*I is in R is a fractional ideal, call it S; observe that for every *i* in I, *i*S, the set of all elements of the form *is*, is contained in R.

Now, SI is obviously an integral ideal of R. But it’s not necessarily the whole of R; that needs proof. I’m going to prove that tomorrow, by first proving it in the case I is a prime (integral) ideal of R, and then using that to prove unique factorization into ideals. Then if I = (P1^*a*1)(P2^*a*2)…(P(n)^*a*(n)), then S = (P1^(-*a*1))(P2^(-*a*2))…(P(n)^(-*a*(n))) and we’re done.

[...] Second, continuing from my previous post about ideal theory, I’m going to show that given any fractional ideal I of a Dedekind domain, there exists another fractional ideal J such that IJ is precisely the domain. [...]

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